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\begin{center}
\vskip 1cm{\LARGE\bf On a Family of Generalized Pascal Triangles
\\
\vskip .1in
Defined by Exponential Riordan Arrays} \vskip 1cm \large
Paul Barry\\
School of Science\\
Waterford Institute of Technology\\
Ireland\\
\href{mailto:pbarry@wit.ie}{\tt pbarry@wit.ie} \\
\end{center}
\vskip .2 in

\begin{abstract} 
We introduce a family of number triangles defined by exponential
Riordan arrays, which generalize Pascal's triangle. We characterize
the row sums and central coefficients of these triangles, and define
and study a set of generalized Catalan numbers. We establish links
to the Hermite, Laguerre and Bessel polynomials, as well as links to
the Narayana and Lah numbers.
\end{abstract}

\section{Introduction}
In \cite{PasTri}, we studied a family of generalized Pascal triangles whose elements
were defined by Riordan arrays, in the sense of \cite{SGWW,Spru}. In this note,
we use so-called ``exponential Riordan arrays'' to define another family of generalized
Pascal triangles. These number triangles are easy to describe, and important number sequences
derived from them are linked to both the Hermite and Laguerre polynomials, as well as
being related to the Narayana and Lah numbers.

We begin by looking at Pascal's triangle, the binomial transform, exponential Riordan arrays,
the Narayana numbers, and
briefly summarize those features of the Hermite and Laguerre polynomials that we will require.
We then introduce the family of generalized Pascal triangles based on exponential Riordan arrays, and
look at a simple case in depth. We finish by enunciating a set of general results concerning row sums,
central coefficients and generalized Catalan numbers for these triangles.

\section{Preliminaries}
Pascal's triangle, with general term $C(n,k)=\binom{n}{k}$, $n,k\geq 0$,
has fascinated mathematicians by its wealth of properties since its
discovery \cite{Pas}. Viewed as an infinite lower-triangular matrix, it is
invertible, with an inverse whose general term is given by
$(-1)^{n-k}\binom{n}{k}$. Invertibility follows from the fact that
$\binom{n}{n}=1$. It is \emph{centrally symmetric}, since by definition,
$\binom{n}{k}=\binom{n}{n-k}$. All the terms of this matrix are
integers.

By a generalized Pascal triangle we shall understand a
lower-triangular infinite integer matrix $T=T(n,k)$ with
$T(n,0)=T(n,n)=1$ and $T(n,k)=T(n,n-k)$. We index all matrices in
this paper beginning at the $(0,0)$-th element.

We shall encounter transformations that operate on integer sequences
during the course of this note. An example of such a transformation
that is widely used in the study of integer sequences is the
so-called Binomial transform \cite{Bin}, which associates to the
sequence with general term $a_n$ the sequence with general term
$b_n$ where
\begin{equation}b_n=\sum_{k=0}^n{{n}\choose{k}}a_k.\end{equation}
If we consider the sequence with general term $a_n$ to be the vector
$\mathbf{a}=(a_0,a_1,\ldots)$
 then we obtain the binomial transform of the sequence by multiplying this (infinite)
 vector by the lower-triangle matrix $\mathbf{B}$ whose $(n,k)$-th element is equal to
$\binom{n}{k}$:
\begin{displaymath}\mathbf{B}=\left(\begin{array}{ccccccc} 1 & 0 & 0
& 0 & 0 & 0 & \ldots \\1 & 1 & 0 & 0 & 0 & 0 & \ldots \\ 1 & 2 & 1
& 0 & 0 & 0 & \ldots \\ 1 & 3 & 3 & 1 & 0 & 0 & \ldots \\ 1 & 4 &
6 & 4 & 1 & 0 & \ldots
\\1 & 5 & 10 & 10 & 5 & 1 &\ldots\\ \vdots & \vdots & \vdots & \vdots & \vdots
& \vdots & \ddots\end{array}\right)\end{displaymath} This
transformation is invertible, with
\begin{equation}a_n=\sum_{k=0}^n
{{n}\choose{k}}(-1)^{n-k}b_k.\end{equation} We note that
$\mathbf{B}$ corresponds to Pascal's triangle. Its row sums are
$2^n$, while its diagonal sums are the Fibonacci numbers $F(n+1)$.
If $\mathbf{B}^m$ denotes the $m-$th power of $\mathbf{B}$, then the
$n-$th term of $\mathbf{B}^m\mathbf{a}$ where $\mathbf{a}=\{a_n\}$
is given by $\sum_{k=0}^n m^{n-k}{{n}\choose{k}}a_k$.

If $\mathcal{A}(x)$ is the ordinary generating function of the
sequence $a_n$, then the ordinary generating function of the
transformed sequence $b_n$ is
$\frac{1}{1-x}\mathcal{A}(\frac{x}{1-x})$. Similarly, if
$\mathcal{G}(x)$ is the exponential generating function (e.g.f.) of
the sequence $a_n$, then the exponential generating function of the
binomial transform of $a_n$ is $\exp(x)\mathcal{G}(x)$.

 The binomial transform is an element of the exponential Riordan
group, which can be defined as follows.

The \emph{exponential Riordan group} \cite{DeutschShap}, is a set of
infinite lower-triangular integer matrices, where each matrix is
defined by a pair of generating functions
$g(x)=1+g_1x+g_2x^2+\ldots$ and $f(x)=f_1x+f_2x^2+\ldots$ where
$f_1\ne 0$. The associated matrix is the matrix whose
$k$-th column has exponential generating function $g(x)f(x)^k/k!$ (the
first column being indexed by 0). The matrix corresponding to the
pair $f, g$ is denoted by $(g, f)$ or $\cal{R}$$(g,f)$. The group
law is then given by
\begin{displaymath} (g, f)*(h, l)=(g(h\circ f), l\circ
f).\end{displaymath} The identity for this law is $I=(1,x)$ and the
inverse of $(g, f)$ is $(g, f)^{-1}=(1/(g\circ \bar{f}), \bar{f})$
where $\bar{f}$ is the compositional inverse of $f$.

If $\mathbf{M}$ is the matrix $(g,f)$, and $\mathbf{a}=\{a_n\}$ is
an integer sequence with exponential generating function
$\mathcal{A}$ $(x)$, then the sequence $\mathbf{M}\mathbf{a}$ has
exponential generating function $g(x)\mathcal{A}(f(x))$.

We note at this juncture that the exponential Riordan group, as well as the group of
`standard' Riordan arrays \cite{SGWW} can be cast in the more general context of matrices of type
$R^q(\alpha_n, \beta_k; \phi,f,\psi)$ as found in \cite{Egor,EZDecomp,EZInt}.
Specifically,
a matrix $C=(c_{nk})_{n,k=0,1,2,\ldots}$ is of type $R^q(\alpha_n, \beta_k; \phi,f,\psi)$ if its general
term is defined by the formula
$$c_{nk}=\frac{\beta_k}{\alpha_n}\mathbf{res}_x(\phi(x)f^k(x)\psi^n(x)x^{-n+qk-1})$$
where $\mathbf{res}_xA(x)=a_{-1}$ for a given formal power series $A(x)=\sum_j a_jx^j$ is the formal residue of
the series.

For the exponential Riordan arrays in this note, we have $\alpha_n=\frac{1}{n!}$, $\beta_k=\frac{1}{k!}$, and
$q=1$.

\begin{example}  The Binomial matrix $\mathbf{B}$ is the element $(e^x,x)$
of the exponential Riordan group. More generally, $\mathbf{B}^m$ is
the element $(e^{mx},x)$ of the Riordan group. It is easy to show
that the inverse $\mathbf{B}^{-m}$ of $\mathbf{B}^m$ is given by
$(e^{-mx},x)$.
\end{example}
\begin{example}
The exponential generating function of the row sums of the matrix
$(g, f)$ is obtained by applying $(g, f)$ to $e^x$, the e.g.f. of
the sequence $1,1,1,\ldots$. Hence the row sums of $(g,f)$ have
e.g.f. $g(x)e^{f(x)}$.
\end{example}

We shall frequently refer to sequences by their sequence number in
the On-Line Encylopedia of Integer Sequences \cite{SL1,SL2}.
For instance, Pascal's triangle is \seqnum{A007318} while the
Catalan numbers \cite{Cat} $C(n)=\binom{2n}{n}/(n+1)$ are \seqnum{A000108}.

\begin{example} An example of a well-known centrally symmetric
invertible triangle is the Narayana triangle $\mathbf{\tilde{N}}$,
\cite{SulankeNara1,SulankeNara2}, defined by
\begin{displaymath}
\tilde{N}(n,k)=\frac{1}{k+1}\binom{n}{k}\binom{n+1}{k}=\frac{1}{n+1}\binom{n+1}{k+1}\binom{n+1}{k}\end{displaymath}
for $n,k\ge 0$.
Other expressions for $\tilde{N}(n,k)$ are given by
\begin{displaymath}
\tilde{N}(n,k)={\binom{n}{k}}^2-\binom{n}{k+1}\binom{n}{k-1}=\binom{n+1}{k+1}\binom{n}{k}
-\binom{n+1}{k}\binom{n}{k+1}.\end{displaymath}

This triangle begins
\begin{displaymath}\mathbf{\tilde{N}}=\left(\begin{array}{ccccccc} 1 & 0 & 0
& 0 & 0 & 0 & \ldots \\1 & 1 & 0 & 0 & 0 & 0 & \ldots \\ 1 & 3 & 1 &
0 & 0 & 0 & \ldots \\ 1 & 6 & 6 & 1 & 0 & 0 & \ldots \\ 1 & 10 & 20
& 10 & 1 & 0 & \ldots
\\1 & 15 & 50 & 50 & 15 & 1 &\ldots\\ \vdots & \vdots & \vdots & \vdots & \vdots
& \vdots & \ddots\end{array}\right)\end{displaymath} Note that in
the literature, it is often the triangle
$\tilde{N}(n-1,k-1)=\frac{1}{n}\binom{n}{k}\binom{n}{k-1}$ that is
referred to as the Narayana triangle. Alternatively, the triangle
$\tilde{N}(n-1,k)=\frac{1}{k+1}\binom{n-1}{k}\binom{n}{k}$ is
referred to as the Narayana triangle. We shall denote this latter
triangle by $N(n,k)$. We then have
\begin{displaymath}\mathbf{\mathbf{N}}=\left(\begin{array}{ccccccc} 1 & 0 & 0
& 0 & 0 & 0 & \ldots \\1 & 0 & 0 & 0 & 0 & 0 & \ldots \\ 1 & 1 & 0 &
0 & 0 & 0 & \ldots \\ 1 & 3 & 1 & 0 & 0 & 0 & \ldots \\ 1 & 6 & 6 &
1 & 0 & 0 & \ldots
\\1 & 10 & 20 & 10 & 1 & 0 &\ldots\\ \vdots & \vdots & \vdots & \vdots & \vdots
& \vdots & \ddots\end{array}\right)\end{displaymath} with row sums equal to the Catalan numbers $C(n)$.

Note that for $n,k\ge 1$, $N(n,k)=\frac{1}{n}\binom{n}{k}\binom{n}{k+1}$. We have,
for instance,
\begin{displaymath}\tilde{N}(n-1,k-1)=\frac{1}{n}\binom{n}{k}\binom{n}{k-1}={\binom{n}{k}}^2-\binom{n-1}{k}\binom{n+1}{k}=
\binom{n}{k}\binom{n-1}{k-1}-\binom{n}{k-1}\binom{n-1}{k}.\end{displaymath}
The last expression represents a $2\times2$ determinant of adjacent
elements in Pascal's triangle. The Narayana triangle is
\seqnum{A001263}.
\end{example}
The Hermite polynomials $H_n(x)$ \cite{Hermite} are defined by
$$H_n(x)=(-1)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2}.$$
They obey $H_n(-x)=(-1)^nH_n(x)$ and can be defined by the
recurrence
\begin{equation}\label{HerRec} H_{n+1}(x)=2xH_n(x)-2nH_{n-1}(x).\end{equation} They have
a generating function given by
$$e^{2tx-x^2}=\sum_{n=0}^{\infty}\frac{H_n(t)}{n!}x^n.$$
A property that is related to the binomial transform is the
following:
$$\sum_{k=0}^n \binom{n}{k}H_k(x)(2z)^{n-k}=H_n(x+z).$$
From this, we can deduce the following proposition.
\begin{proposition}\label{HerBin} For fixed $x$ and $y\ne 0$, the binomial
transform of the sequence $n \to H_n(x)y^n$ is the sequence $n \to
y^nH_n(x+\frac{1}{2y})$. \end{proposition}
\begin{proof} Let $z=\frac{1}{2y}$. Then $2z=\frac{1}{y}$ and hence
$$\sum_{k=0}^n \binom{n}{k}H_k(x)(y)^{k-n}=H_n(x+\frac{1}{2y}).$$
That is,
$$\sum_{k=0}^n \binom{n}{k}H_k(x)y^k=y^nH_n(x+\frac{1}{2y})$$
as required.
\end{proof}

The Laguerre polynomials $L_n(x)$ \cite{Laguerre} are defined by
$$L_n(x)=\frac{e^x}{n!}\frac{d^n}{dx^n}x^ne^{-x}.$$
They have generating function
$$\frac{\exp(-\frac{tx}{1-x})}{1-x}=\sum_{n=0}^{\infty}\frac{L_n(t)}{n!}x^n.$$
They are governed by the following recurrence relationship:
\begin{equation}\label{LagRec}
(n+1)L_{n+1}(t)=(2n+1-t)L_n(t)-nL_{n-1}(t)\end{equation}

\section{Introducing the family of centrally symmetric invertible triangles}
We recall that the Binomial matrix $\mathbf{B}$, or Pascal's
triangle, is the element $(e^x, x)$ of the Riordan group. For a
given integer $r$, we shall denote by $\mathbf{B}_r$ the element
$(e^x, x(1+rx))$ of the Riordan group. We note that
$\mathbf{B}=\mathbf{B}_0$. We can characterize the general element
of $\mathbf{B}_r$ as follows.
\begin{proposition} The general term $B_r(n,k)$ of the matrix
$\mathbf{B}_r$ is given by
$$B_r(n,k)=\frac{n!}{k!}\sum_{j=0}^k
\binom{k}{j}\frac{r^j}{(n-k-j)!}.$$
\end{proposition}
\begin{proof} We have
\begin{eqnarray*}B_r(n,k)&=&\frac{n!}{k!}[x^n](e^x(x(1+rx)^k)\\
&=&\frac{n!}{k!}[x^n]\sum_{i=0}^{\infty}\frac{x^i}{i!}x^k\sum_{j=0}^k\binom{k}{j}r^jx^j\\
&=&\frac{n!}{k!}[x^{n-k}]\sum_{i=0}^{\infty}\sum_{j=0}^k\binom{k}{j}\frac{r^j}{i!}x^{i+j}\\
&=&\frac{n!}{k!}\sum_{j=0}^k\binom{k}{j}\frac{r^j}{(n-k-j)!}.\end{eqnarray*}
\end{proof}
From the above expression we can easily establish that $B_r(n,k)=B_r(n,n-k)$ and $B_r(n,0)=B_r(n,n)=1$.

An alternative derivation of these results can be obtained be observing that
the matrix $\mathbf{B}_r$ may be defined as the array $R^1(\frac{1}{n!},\frac{1}{k!};e^x,(1+rx),1)$.
Then we have
\begin{eqnarray*} B_r(n,k)&=&\frac{1/k!}{1/n!}\mathbf{res}_x(e^x(1+rx)^kx^{-n+k-1})\\
&=&\frac{n!}{k!}\mathbf{res}_x(\sum_i^{\infty}\frac{x^i}{i!}\sum_j^k \binom{k}{j}r^jx^jx^{-n+k-1})\\
&=&\frac{n!}{k!}\mathbf{res}_x(\sum_i^{\infty}\sum_j^k \binom{k}{j}\frac{r^j}{i!}x^{i+j-n+k-1})\\
&=&\frac{n!}{k!}\sum_j^k \binom{k}{j}\frac{r^j}{(n-k-j)!}.\end{eqnarray*}
Thus $\mathbf{B_r}$ is a centrally symmetric lower-triangular matrix
with $B_r(n,0)=B_r(n,n)=1$. In this sense $\mathbf{B}_r$ can be regarded as a
generalized Pascal matrix. Note that by the last property, this
matrix is invertible.
\begin{proposition} The inverse of $\mathbf{B_r}$ is the element
$(e^{-u}, u)$ of the Riordan group, where
$$u=\frac{\sqrt{1+4rx}-1}{2r}.$$
\end{proposition}
\begin{proof} Let $(g^*,\bar{f})$ be the inverse of $(e^x,
x(1+rx))$. Then
$$(g^*,\bar{f})(e^x,x(1+rx))=(1,x)\Rightarrow
\bar{f}(1+r\bar{f})=x.$$ Solving for $\bar{f}$ we get
$$\bar{f}=\frac{\sqrt{1+4rx}-1}{2r}.$$
But $g^*=\frac{1}{g\circ\bar{f}}=e^{-\bar{f}}$.
\end{proof}
This result allows us to easily characterize the row sums of the inverse $\mathbf{B}_r^{-1}$.
\begin{corollary} The row sums of the inverse triangle $\mathbf{B}_r^{-1}$ are given by
$0^n=1,0,0,0,\ldots$.
\end{corollary}
\begin{proof} We have $\mathbf{B}_r^{-1}=(e^{-u},u)$ as above. Hence the e.g.f. of the
row sums of $\mathbf{B}_r^{-1}$ is $e^{-u}e^u=1$. The result follows from this. \end{proof}
\begin{example}\label{ExB1} $\mathbf{B}_1=(e^x, x(1+x))$ is given by
\begin{displaymath}\mathbf{B}_1=\left(\begin{array}{ccccccc} 1 & 0 & 0
& 0 & 0 & 0 & \ldots \\1 & 1 & 0 & 0 & 0 & 0 & \ldots \\ 1 & 4 & 1 &
0 & 0 & 0 & \ldots \\ 1 & 9 & 9 & 1 & 0 & 0 & \ldots \\ 1 & 16 & 42
& 16 & 1 & 0 & \ldots
\\1 & 25 & 130 & 130 & 25 & 1 &\ldots\\ \vdots & \vdots & \vdots & \vdots & \vdots
& \vdots & \ddots\end{array}\right)\end{displaymath}
\end{example}
The row sums of $\mathbf{B}_1$ are
$$1, 2, 6, 20, 76, 312, 1384, 6512, 32400, \ldots$$ or \seqnum{A000898}.

From the above, the terms of this sequence are given by
$$s_1(n)=\sum_{k=0}^n\frac{n!}{k!}\sum_{j=0}^k
\binom{k}{j}\frac{1}{(n-k-j)!}.$$ with e.g.f.
$g(x)e^{f(x)}=e^xe^{x(1+x)}=e^{2x+x^2}$. What is less evident is
that
$$s_1(n)=H_n(-i)i^n$$ where $i=\sqrt{-1}$. This follows since
\begin{eqnarray*}e^{2x+x^2}&=&e^{2(-i)(ix)-(ix)^2}\\
&=&\sum_{n=0} \frac{H_n(-i)}{n!}(ix)^n\\
&=&\sum_{n=0} \frac{H_n(-i)i^n}{n!}x^n\end{eqnarray*} and hence
$e^{2x+x^2}$ is the e.g.f. of $H_n(-i)i^n$. We therefore obtain the
identity
$$ H_n(-i)i^n=\sum_{k=0}^n\frac{n!}{k!}\sum_{j=0}^k
\binom{k}{j}\frac{1}{(n-k-j)!}.$$

We can characterize the row sums of $\mathbf{B}_1$ in terms of the
diagonal sums of another related special matrix. For this, we recall 
\cite{Bessel}
that $${\rm Bessel}(n,k)=\frac{(n+k)!}{2^k(n-k)!k!}=\binom{n+k}{2}(2k-1)!!$$ defines the triangle
\seqnum{A001498} of coefficients of Bessel polynomials that begins
\begin{displaymath}\mathbf{Bessel}=\left(\begin{array}{ccccccc} 1 & 0 & 0
& 0 & 0 & 0 & \ldots \\1 & 1 & 0 & 0 & 0 & 0 & \ldots \\ 1 & 3 & 3 &
0 & 0 & 0 & \ldots \\ 1 & 6 & 15 & 15 & 0 & 0 & \ldots \\ 1 & 10 &
45 & 105 & 105 & 0 & \ldots
\\1 & 15 & 105 & 420 & 945 & 945 &\ldots\\ \vdots & \vdots & \vdots & \vdots & \vdots
& \vdots & \ddots\end{array}\right)\end{displaymath} We then have
\begin{proposition} The row sums of the matrix $\mathbf{B}_1$ are
equal to the diagonal sums of the matrix with general term
${\rm Bessel}(n,k)2^n$. That is
$$H_n(-i)i^n=\sum_{k=0}^n\frac{n!}{k!}\sum_{j=0}^k
\binom{k}{j}\frac{1}{(n-k-j)!}=\sum_{k=0}^{\lfloor {\frac{n}{2}}
\rfloor} {\rm Bessel}(n-k,k)2^{n-k}.$$
\end{proposition}
\begin{proof} We shall prove this in two steps. First, we shall show
that
$$\sum_{k=0}^{\lfloor {\frac{n}{2}}
\rfloor} {\rm Bessel}(n-k,k)2^{n-k}=\sum_{k=0}^{\lfloor {\frac{n}{2}}
\rfloor}\frac{(2k)!}{k!}\binom{n}{2k}2^{n-2k}=\sum_{k=0}^{\lfloor
{\frac{n}{2}} \rfloor}(2k-1)!!\binom{n}{2k}2^{n-k}.$$ We shall then
show that this is equal to $H_n(-i)i^n$. Now
\begin{eqnarray*}\sum_{k=0}^{\lfloor {\frac{n}{2}}
\rfloor}
{\rm Bessel}(n-k,k)2^{n-k}&=&\sum_{k=0}^n {\rm Bessel}(n-\frac{k}{2},\frac{k}{2})2^{n-\frac{k}{2}}(1+(-1)^k)/2\\
&=&\sum_{k=0}^n\frac{(n-\frac{k}{2}+\frac{k}{2})!2^{n-\frac{k}{2}}}{2^{\frac{k}{2}}(n-\frac{k}{2}-\frac{k}{2})!(\frac{k}{2})!}(1+(-1)^k)/2\\
&=&\sum_{k=0}^n\frac{n!}{(n-k)!}\frac{2^{n-k}}{(\frac{k}{2})!}(1+(-1)^k)/2\\
&=&\sum_{k=0}^n\frac{k!}{(\frac{k}{2})!}\binom{n}{k}2^{n-k}(1+(-1)^k)/2\\
&=&\sum_{k=0}^{\lfloor {\frac{n}{2}}
\rfloor}\frac{(2k)!}{k!}\binom{n}{2k}2^{n-2k}\\
&=&\sum_{k=0}^{\lfloor {\frac{n}{2}}
\rfloor}\frac{(2k)!}{2^k k!}\binom{n}{2k}2^{n-k}\\
&=&\sum_{k=0}^{\lfloor {\frac{n}{2}}
\rfloor}(2k-1)!!\binom{n}{2k}2^{n-k}.
\end{eqnarray*} establishes the first part of the proof.
The second part of the proof is a consequence of the following more
general result, when we set $a=2$ and $b=1$.
\end{proof}
\begin{proposition} The sequence with e.g.f. $e^{ax+bx^2}$ has
general term $u_n$ given by
$$ u_n=\sum_{k=0}^{\lfloor {\frac{n}{2}}
\rfloor}\binom{n}{2k}\frac{(2k)!}{k!}a^{n-2k}b^k=\sum_{k=0}^{\lfloor
{\frac{n}{2}} \rfloor}\binom{n}{2k}C(k)(k+1)!a^{n-2k}b^k.$$
\end{proposition}
\begin{proof} We have
\begin{eqnarray*}n![x^n]e^{ax+bx^2}&=&n![x^n]e^{ax}e^{bx^2}\\
&=&n![x^n]\sum_{i=0}^{\infty}\frac{a^ix^i}{i!}\sum_{k=0}^{\infty}\frac{b^kx^{2k}}{k!}\\
&=&n![x^n]\sum_{i=0}^{\infty}\sum_{k=0}^{\infty}\frac{a^ib^k}{i!k!}x^{i+2k}\\
&=&n!\sum_{k=0}^{\infty}\frac{a^{n-2k}b^k}{(n-2k)!k!}\\
&=&\sum_{k=0}^{\infty}\frac{n!}{(n-2k)!(2k)!}\frac{(2k)!}{k!}a^{n-2k}b^k\\
&=&\sum_{k=0}^{\lfloor {\frac{n}{2}}
\rfloor}\binom{n}{2k}\frac{(2k)!}{k!}a^{n-2k}b^k.
\end{eqnarray*}
\end{proof}
\begin{corollary} $$H_n(-\frac{a}{2\sqrt{b}}i)(\sqrt{b}i)^n=\sum_{k=0}^{\lfloor {\frac{n}{2}}
\rfloor}\binom{n}{2k}\frac{(2k)!}{k!}a^{n-2k}b^k.$$
\end{corollary}
\begin{corollary} Let $u_n$ be the sequence with e.g.f.
$e^{ax+bx^2}$. Then $u_n$ satisfies the recurrence $$
u_n=au_{n-1}+2(n-1)bu_{n-2}$$ with $u_0=1$, $u_1=a$.
\end{corollary}
\begin{proof} Equation \ref{HerRec} implies that
$$H_{n}(x)=2xH_{n-1}(x)-2(n-1)H_{n-2}(x).$$
Thus
$$H_n(-\frac{a}{2\sqrt{b}}i)=-2\frac{a}{2\sqrt{b}}iH_{n-1}(-\frac{a}{2\sqrt{b}}i)-2(n-1)H_{n-2}(-\frac{a}{2\sqrt{b}}i).$$
Now multiply both sides by $(\sqrt{b}i)^n$ to obtain
$$
u_n=au_{n-1}+2(n-1)bu_{n-2}.$$ Since
$$u_n=\sum_{k=0}^{\lfloor {\frac{n}{2}}
\rfloor}\binom{n}{2k}\frac{(2k)!}{k!}a^{n-2k}b^k$$ we obtain the initial values
 $u_0=1$, $u_1=a$.
\end{proof}
\begin{corollary}
The binomial transform of $\sum_{k=0}^{\lfloor {\frac{n}{2}}
\rfloor}\binom{n}{2k}\frac{(2k)!}{k!}a^{n-2k}b^k$ is given by
$$\sum_{k=0}^{\lfloor {\frac{n}{2}}
\rfloor}\binom{n}{2k}\frac{(2k)!}{k!}(a+1)^{n-2k}b^k. $$
\end{corollary}
\begin{proof}
The e.g.f. of the binomial transform of the sequence with e.g.f.
$e^{ax+cx^2}$ is $e^xe^{ax+bx^2}=e^{(a+1)x+bx^2}$.
\end{proof}
Equivalently, the binomial transform of $\sum_{k=0}^{\lfloor
{\frac{n}{2}} \rfloor}\binom{n}{2k}C(k)(k+1)!a^{n-2k}b^k$ is given
by $$\sum_{k=0}^{\lfloor {\frac{n}{2}}
\rfloor}\binom{n}{2k}C(k)(k+1)!(a+1)^{n-2k}b^k.$$ We note that in
\cite{PasTri}, it was shown that the binomial transform of
$\sum_{k=0}^{\lfloor {\frac{n}{2}}
\rfloor}\binom{n}{2k}C(k)a^{n-2k}b^k$ is given by
$$\sum_{k=0}^{\lfloor {\frac{n}{2}}
\rfloor}\binom{n}{2k}C(k)(a+1)^{n-2k}b^k.$$
\begin{corollary} The row sums of $\mathbf{B}_1$ satisfy the recurrence
equation
$$ u_n=2u_{n-1}+2(n-1)u_{n-2} $$ with $u_0=1$, $u_1=2$.
\end{corollary}
 We can use \textbf{Proposition} \ref{HerBin} to study
the inverse binomial transform of $s_1(n)$. By that proposition, the
inverse binomial transform of $H_n(-i)i^n$ is given by
$i^nH_n(-i+\frac{1}{2i})=H_n(-\frac{i}{2})i^n$. This is the sequence
$$1, 1, 3, 7, 25, 81, 331, 1303, 5937, \ldots$$ with e.g.f. $e^{x+x^2}$.
This is \seqnum{A047974} which satisfies the recurrence $a_n =
a_{n-1}+2(n-1)a_{n-2}$. It is in fact equal to $\sum_{k=0}^{\lfloor
{\frac{n}{2}} \rfloor} {\rm Bessel}(n-k,k)2^k$. The second inverse
binomial transform of $s_1(n)$ is the sequence
$$1, 0, 2, 0, 12, 0, 120, 0, 1680, 0, 30240,\ldots$$ with e.g.f. $e^{x^2}$.
This is an ``aerated'' version of the quadruple factorial numbers $C(n)(n+1)!=\frac{(2n)!}{n!}$, or
\seqnum{A001813}.

We now look at the central coefficients $B_1(2n,n)$ of $\mathbf{B}_1$. We have
\begin{eqnarray*}B_1(2n,n)&=&\frac{(2n)!}{n!}\sum_{j=0}^n \binom{n}{j}\frac{1}{(n-j)!}\\
&=&C(n)(n+1)!\sum_{j=0}^n\binom{n}{j}\frac{1}{(n-j)!}\\
&=&C(n)(n+1)\sum_{j=0}^n\binom{n}{j}^2 j!\\
&=&C(n)(n+1)!L_n(-1).\end{eqnarray*} Hence $$
\frac{B_1(2n,n)}{C(n)(n+1)!}=L_n(-1).$$ We note that this is the
rational sequence
$1,2,\frac{7}{2},\frac{17}{3},\frac{209}{24},\ldots$.  Two other
ratios are of interest.
\begin{enumerate}
\item $\frac{B_1(2n,n)}{C(2n,n)}=n!L_n(-1)$ is \seqnum{A002720}. It has e.g.f. $\frac{1}{1-x}\exp(\frac{x}{1-x})$. It is equal
to the number of partial permutations of an $n$-set, as well as the number of matchings in the bipartite graph $K(n,n)$.
Using Equation (\ref{LagRec}) we can show that these numbers obey the following recurrence:
$$u_n=2nu_{n-1}-(n-1)^2u_{n-2}$$
with $u_0=1$, $u_1=2$.
\item $\frac{B_1(2n,n)}{C(n)}=(n+1)!L_n(-1)$ is \seqnum{A052852}$(n+1)$. It has e.g.f. given by
$$\frac{d}{dx}\frac{x}{1-x}\exp(\frac{x}{1-x})=\frac{1}{(1-x)^3}\exp(\frac{x}{1-x}).$$
Again using Equation (\ref{LagRec}) we can show that these numbers obey the following recurrence:
 $$v_n=2(n+1)v_{n-1}-(n^2-1)v_{n-2}$$ with $v_0=1$, $v_1=4$.

 This sequence counts
 the number of $(121,212)$-avoiding $n$-ary words of length $n$. Specifically,
 $$\frac{B_1(2n,n)}{C(n)}=f_{121,212}(n+1,n+1)$$ where
 $$f_{121,212}(n,k)=\sum_{j=0}^k\binom{k}{j}\binom{n-1}{j-1}j!$$ is defined in \cite{Burstein}.

\end{enumerate}
From this last point, we find the following expression
\begin{equation}\label{avoid}B_1(2n,n)=C(n)\sum_{j=0}^{n+1}\binom{n+1}{j}\binom{n}{j-1}j!\end{equation}
Based on the fact that $C(n)=\binom{2n}{n}-\binom{2n}{n-1}$ we define $$
C_1(n)=B_1(2n,n)-B_1(2n,n-1)=B_1(2n,n)-B_1(2n,n+1)$$ to be \emph{the
generalized Catalan numbers associated with the triangle} $\mathbf{B}_1$.
We calculate $B_1(2n,n-1)$ as follows:
\begin{eqnarray*} B_1(2n,n-1)&=&\frac{(2n)!}{(n-1)!}\sum_{j=0}^{n-1}\binom{n-1}{j}\frac{1}{(n-j+1)!}\\
&=&\frac{(2n)!}{(n-1)!}\sum_{j=0}^{n-1}\binom{n}{j}\frac{n-j}{n}\frac{1}{(n-j+1)!}\\
&=&\frac{(2n)!}{n!}\sum_{j=0}^{n}\binom{n}{j}\frac{1}{(n-j)!}\frac{n-j}{n-j+1}.\end{eqnarray*}
Hence
\begin{eqnarray*}B_1(2n,n)-B_1(2n,n-1)&=&\frac{(2n)!}{n!}\sum_{j=0}^{n}\binom{n}{j}\frac{1}{(n-j)!}(1-\frac{n-j}{n-j+1})\\
&=&\frac{(2n)!}{n!}\sum_{j=0}^{n}\binom{n}{j}\frac{1}{(n-j)!}\frac{1}{n-j+1}\\
&=&\frac{(2n)!}{n!}\sum_{j=0}^{n}\binom{n}{j}\frac{1}{(n-j+1)!}.\end{eqnarray*}
Starting from the above, we can find many expressions for $C_1(n)$. For example,
\begin{eqnarray*}C_1(n)&=&\frac{(2n)!}{n!}\sum_{j=0}^{n}\binom{n}{j}\frac{1}{(n-j+1)!}\\
&=&C(n)\sum_{j=0}^n\binom{n}{j}\frac{(n+1)!}{(n+1-j)!}\\
&=&C(n)\sum_{j=0}^n\binom{n}{j}\binom{n+1}{j}j!\\
&=&C(n)\sum_{j=0}^n\binom{n}{j}^2\frac{n+1}{n-j+1}j!\\
&=&C(n)\sum_{j=0}^n\binom{n}{j}\binom{n+1}{j+1}\frac{(j+1)!}{n-j+1}.\end{eqnarray*}
where we have used the fact that $\frac{(2n)!}{n!}=C(n)(n+1)!$. This is the sequence
\seqnum{A001813} of quadruple factorial numbers with e.g.f. $\frac{1}{\sqrt{1-4x}}$.

Recognizing that the terms after $C(n)$ represent convolutions, we can also write
\begin{eqnarray*}C_1(n)&=&C(n)\sum_{j=0}^n\binom{n}{j}\frac{(n+1)!}{(j+1)!}\\
&=&C(n)\sum_{j=0}^n\binom{n}{j}\binom{n+1}{j+1}(n-j)!\\
&=&C(n)\sum_{j=0}^n\binom{n}{j}^2\frac{n+1}{j+1}(n-j)!\end{eqnarray*}
 We note that the first expression
immediately above links $C_1(n)$ to the Lah numbers \seqnum{A008297}.

The ratio $\frac{C_1(n)}{C(n)}$, or $\sum_{j=0}^n\binom{n}{j}\frac{(n+1)!}{(k+1)!}$, is the sequence
$$1, 3, 13, 73, 501, 4051, 37633, 394353, 4596553, \ldots$$ or \seqnum{A000262}($n+1$). This is related to the
number of partitions of $[n]=\{1,2,3,\ldots,n\}$ into any number of
lists, where a list means an ordered subset. It also has
applications in quantum physics \cite{Blasiak1}. The sequence has
e.g.f.
$$\frac{d}{dx}e^{\frac{x}{1-x}}=\frac{e^{\frac{x}{1-x}}}{(1-x)^2}.$$
We can in fact describe this ratio in terms of the Narayana numbers
$\tilde{N}(n,k)$ as follows:
\begin{eqnarray*} \frac{C_1(n)}{C(n)}&=&\sum_{j=0}^n\binom{n}{j}\binom{n+1}{j+1}(n-j)!\\
&=&\sum_{j=0}^n\frac{n-j+1}{n+1}\binom{n+1}{j}\binom{n+1}{j+1}(n-j)!\\
&=&\sum_{j=0}^n\frac{1}{n+1}\binom{n+1}{j}\binom{n+1}{j+1}(n-j+1)!\\
&=&\sum_{j=0}^n\tilde{N}(n,j)(n-j+1)!\\
&=&\sum_{j=0}^n\tilde{N}(n,n-j)(j+1)!\\
&=&\sum_{j=0}^n\tilde{N}(n,j)(j+1)!\end{eqnarray*}
Hence we have
$$\frac{C_1(n)}{C(n)}=\sum_{j=0}^n\tilde{N}(n,j)(n-j+1)!=\sum_{j=0}^n\tilde{N}(n,j)(j+1)!=\sum_{j=0}^n\binom{n}{j}\frac{(n+1)!}{(j+1)!}$$
\section{The General Case}
We shall now look at the row sums, central coefficients and
generalized Catalan numbers associated with the general matrix
$\mathbf{B}_r$. In what follows, proofs follow the methods developed
in the last section.
\begin{proposition} The row sums $s_r(n)$ of $\mathbf{B}_r$ are given by $H_n(-\frac{i}{\sqrt{r}})(\sqrt{r}i)^n$.
\end{proposition}
\begin{proof} The row sums of $\mathbf{B}_r$ are given by the sequence
$$\sum_{k=0}^n\frac{n!}{k!}\sum_{j=0}^k\binom{k}{j}\frac{r^j}{(n-k-j)!}$$ with e.g.f.
$g(x)e^{f(x)}=e^xe^{x(1+rx)}=e^{2x+rx^2}$. Now
\begin{eqnarray*}e^{2x+rx^2}&=&e^{2(\frac{-i}{\sqrt{r}})(i\sqrt{r}x)-(i\sqrt{r}x)^2}\\
&=&\sum_{n=0}^{\infty}\frac{H_n(-\frac{i}{\sqrt{r}})}{n!}(i\sqrt{r}x)^n\\
&=&\sum_{n=0}^{\infty}\frac{H_n(-\frac{i}{\sqrt{r}})(i\sqrt{r})^n}{n!}x^n
\end{eqnarray*}
\end{proof}
\begin{corollary}
We have the identity
$$\sum_{k=0}^n\frac{n!}{k!}\sum_{j=0}^k\binom{k}{j}\frac{r^j}{(n-k-j)!}=H_n(-\frac{i}{\sqrt{r}})(\sqrt{r}i)^n=\sum_{k=0}
^{\lfloor {\frac{n}{2}} \rfloor}
\binom{n}{2k}\frac{(2k)!}{k!}2^{n-2k}r^k.$$
\end{corollary} As before, we can
rewrite this using the fact that
$\frac{(2k)!}{k!}=C(k)(k+1)!=2^k(2k-1)!!$.

We note that the second inverse binomial transform of $s_r(n)$ has e.g.f. $e^{rx^2}$.
\begin{proposition} The row sums of $\mathbf{B}_r$ are equal to the
diagonal sums of the matrix with general term ${\rm Bessel}(n,k)2^nr^k$.
That is,
$$\sum_{k=0}^n\frac{n!}{k!}\sum_{j=0}^k\binom{k}{j}\frac{r^j}{(n-k-j)!}=H_n(-\frac{i}{\sqrt{r}})(\sqrt{r}i)^n=\sum_{k=0}^{\lfloor
{\frac{n}{2}}
\rfloor}
{\rm Bessel}(n-k,k)2^{n-k}r^k.$$
\end{proposition}
\begin{proposition} The row sums of $\mathbf{B_r}$ obey the
recurrence
$$u_n=2u_{n-1}+2r(n-1)u_{n-2}$$ with $u_0=1$, $u_1=2$.
\end{proposition}
 We now turn our attention to the central coefficients of
$\mathbf{B}_r$.
\begin{proposition} $B_r(2n,n)=C(n)(n+1)!\sum_{j=0}^n\binom{k}{j}^2j!r^j$
\end{proposition}
\begin{proof} The proof is the same as the calculation for $B_1(2n,n)$ in Example \ref{ExB1}, with the extra factor of $r^j$ to be
taken into account.
\end{proof}
\begin{corollary} $$\frac{B_r(2n,n)}{C(n)(n+1)!}=r^nL_n(-\frac{1}{r})=\sum_{j=0}^n\binom{n}{j}\frac{r^j}{(n-j)!}$$
for $r\ne0$.
\end{corollary}
We note that the above expressions are not integers in general.

For instance, $\frac{B_2(2n,n)}{C(2n,n)}=n!2^nL_n(-\frac{1}{2})$
is \seqnum{A025167}, and
$\frac{B_3(2n,n)}{C(2n,n)}=n!3^nL_n(-\frac{1}{3})$ is
\seqnum{A102757}. In general, we have
\begin{proposition} $\frac{B_r(2n,n)}{C(2n,n)}=n!r^nL_n(-1/r)$ has
e.g.f. $\frac{1}{1-rx}\exp(\frac{x}{1-rx})$, and satisfies the recurrence relation
$$u_n=((2n-1)r+1)u_{n-1}-r^2(n-1)^2 u_{n-2}$$ with $u_0=1$, $u_1=r+1$.
\end{proposition}
\begin{proof} We have
\begin{eqnarray*}n![x^n]\frac{e^{\frac{x}{1-rx}}}{(1-rx)}&=&n![x^n]\sum_{i=0}^{\infty}\frac{1}{i!}\frac{x^i}{(1-rx)^i}(1-rx)^{-1}\\
&=&n![x^n]\sum_{i=0}^{\infty}\frac{1}{i!}x^i(1-rx)^{-i}(1-rx)^{-1}\\
&=&n![x^n]\sum_{i=0}^{\infty}\frac{1}{i!}x^i(1-rx)^{-(i+1)}\\
&=&n![x^n]\sum_{i=0}^{\infty}\frac{1}{i!}x^i\sum_{j=0}\binom{-(i+1)}{j}(-1)^jr^jx^j\\
&=&n![x^n]\sum_{i=0}^{\infty}\frac{1}{i!}\sum_{j=0}\binom{i+j}{j}r^jx^{i+j}\\
&=&n!\sum_{j=0}^n\binom{n}{j}\frac{r^j}{(n-j)!}.
\end{eqnarray*}
To prove the second assertion, we use Equation (\ref{LagRec}) with $t=-\frac{1}{r}$. Multiplying by
$n!r^{n+1}$, we obtain
$$(n+1)!r^{n+1}L_{n+1}(-\frac{1}{r})=(2n+1+\frac{1}{r})r^{n+1}n!L_n(-\frac{1}{r})-r^{n+1}n^2(n-1)!L_{n-1}(-\frac{1}{r}).$$
Simplifying, and letting $n\to n-1$, gives the result.
\end{proof}
\begin{corollary} $\frac{B_r(2n,n)}{C(n)}=(n+1)!r^nL_n(-1/r)$ has
e.g.f. $\frac{d}{dx}\frac{x}{1-rx}\exp(\frac{x}{1-rx})$, and satisfies the recurrence
$$w_n=((2n-1)+r)\frac{n+1}{n}w_{n-1}-r^2(n^2-1)w_{n-2}$$ for $n>1$, with
$w_0=1$ and $w_1=2r+2$.
\end{corollary}
We can generalize Equation (\ref{avoid}) to get
$$B_r(2n,n)=C(n)\sum_{j=0}^{n+1}\binom{n+1}{j}\binom{n}{j-1}j!r^{j-1}.$$
We define \emph{the generalized Catalan numbers associated with the triangles} $\mathbf{B}_r$ to be the numbers
$$C_r(n)=B_r(2n,n)-B_r(2n,n-1).$$ Using the methods of \textbf{Example} \ref{ExB1}, we have
\begin{proposition} We have the following equivalent expressions for $C_r(n)$:
\begin{eqnarray*}C_r(n)&=&\frac{(2n)!}{n!}\sum_{j=0}^n\binom{n}{j}\frac{r^j}{(n-j+1)!}\\
&=&C(n)\sum_{j=0}^n \binom{n}{j}\frac{(n+1)!}{(j+1)!}r^{n-j}\\
&=&C(n)\sum_{j=0}^n \binom{n}{j}\binom{n+1}{j+1}(n-j)!r^{n-j}\\
&=&C(n)\sum_{j=0}^n \frac{n+1}{j+1}\binom{n}{j}^2(n-j)!r^{n-j}\\
&=&C(n)\sum_{j=0}^n \tilde{N}(n,j)(j+1)!r^j.\end{eqnarray*}
\end{proposition}
For instance, $C_2(n)/C(n)$ is \seqnum{A025168}.
\begin{proposition} $\frac{C_r(n)}{C(n)}$ has e.g.f.
$$\frac{d}{dx}e^{\frac{x}{1-rx}}=\frac{e^{\frac{x}{1-rx}}}{(1-rx)^2}.$$
\end{proposition}
\begin{proof} We have
\begin{eqnarray*}n![x^n]\frac{e^{\frac{x}{1-rx}}}{(1-rx)^2}&=&n![x^n]\sum_{i=0}^{\infty}\frac{1}{i!}\frac{x^i}{(1-rx)^i}(1-rx)^{-2}\\
&=&n![x^n]\sum_{i=0}^{\infty}\frac{1}{i!}x^i(1-rx)^{-i}(1-rx)^{-2}\\
&=&n![x^n]\sum_{i=0}^{\infty}\frac{1}{i!}x^i(1-rx)^{-(i+2)}\\
&=&n![x^n]\sum_{i=0}^{\infty}\frac{1}{i!}x^i\sum_{j=0}^{\infty}\binom{-(i+2)}{j}(-1)^jr^jx^j\\
&=&n![x^n]\sum_{i=0}^{\infty}\frac{1}{i!}\sum_{j=0}^{\infty}\binom{i+j+1}{j}r^jx^{i+j}\\
&=&n!\sum_{j=0}^{\infty}\binom{n+1}{j}\frac{r^j}{(n-j)!}\\
&=&(n+1)!\sum_{j=0}^n\binom{n}{j}\frac{r^j}{(n-j+1)!}.
\end{eqnarray*}
\end{proof}
\section{The case $r=\frac{1}{2}$}
The assumption so far has been that $r$ is an integer. In this
section, we indicate that $r=\frac{1}{2}$ also produces a
generalized Pascal triangle. We have
$\mathbf{B}_{\frac{1}{2}}=(e^x,x(1+x/2))$. This begins
\begin{displaymath}\mathbf{B}_{\frac{1}{2}}=\left(\begin{array}{ccccccc} 1 & 0 & 0
& 0 & 0 & 0 & \ldots \\1 & 1 & 0 & 0 & 0 & 0 & \ldots \\ 1 & 3 & 1 &
0 & 0 & 0 & \ldots \\ 1 & 6 & 6 & 1 & 0 & 0 & \ldots \\ 1 & 10 & 21
& 10 & 1 & 0 & \ldots
\\1 & 15 & 55 & 55 & 15 & 1 &\ldots\\ \vdots & \vdots & \vdots & \vdots & \vdots
& \vdots & \ddots\end{array}\right)\end{displaymath} This is
triangle \seqnum{A100862}. Quoting from \seqnum{A100862},
$B_{\frac{1}{2}}(n,k)$ ``is the number of $k$-matchings of the
corona $K'(n)$ of the complete graph $K(n)$ and the complete graph
$K(1)$; in other words, $K'(n)$ is the graph constructed from $K(n)$
by adding for each vertex $v$ a new vertex $v'$ and the edge
$vv'$''. The row sums of this triangle, \seqnum{A005425}, are given
by
$$1, 2, 5, 14, 43, 142, 499, 1850, 7193, \ldots$$ These have e.g.f.
$e^{2x+x^2/2}$ and general term
$$H_n(-\sqrt{2}i)(i/\sqrt{2})^n=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}\binom{n}{2k}\frac{(2k)!}{k!}2^{n-3k}.$$
They obey the recurrence $$ u_n=2u_{n-1}+(n-1)u_{n-2} $$ with
$u_0=1$, $u_1=2$.

 \cite{Blasiak2} provides an example of their use
in quantum physics. Using Proposition \ref{HerBin} or otherwise, we
see that the inverse binomial transform of this sequence, with
e.g.f. $e^{x+x^2/2}$, is given by
$$H_n(-\sqrt{2}i+\frac{i}{\sqrt{2}})(i/\sqrt{2})^n=H_n(-\frac{i}{\sqrt{2}})(i/\sqrt{2})^n.$$ This is the
sequence $$1,1,2,4,10,26,76,232,765,\ldots$$ or \seqnum{A000085}. It
has many combinatorial interpretations, including for instance the
number of matchings in the complete graph $K(n)$. These numbers are
the diagonal sums of the Bessel triangle $\mathbf{Bessel}$:
$$H_n(-\frac{i}{\sqrt{2}})(i/\sqrt{2})^n=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}{\rm Bessel}(n-k,k).$$
The row sums of $\mathbf{B}_{\frac{1}{2}}$ are the second binomial
transform of the sequence
$$1, 0, 1, 0, 3, 0, 15, 0, 105, 0,\ldots$$ with e.g.f. $e^{x^2/2}$.
This is an ``aerated'' version of the double factorial numbers
$(2n-1)!!$, or \seqnum{A001147}. These count the number of perfect
matchings in the complete graph $K(2n)$. The row sums count the
number of $12-3$ and $214-3$-avoiding permutations, as well as the
number of matchings of the corona $K'(n)$ of the complete graph
$K(n)$ and the complete graph $K(1)$.

This example prompts us to define a new family
$\tilde{\mathbf{B}}_r$ where $\tilde{\mathbf{B}}_r$ is the element
$(e^x, x(1+\frac{rx}{2}))$ of the exponential Riordan group. Then we
have $\tilde{\mathbf{B}}_0=\mathbf{B}$,
$\tilde{\mathbf{B}}_1=\mathbf{B}_{\frac{1}{2}}$,
$\tilde{\mathbf{B}}_2=\mathbf{B}_1$ etc. We can then show that
$\tilde{\mathbf{B}}_r$ is the product of the binomial matrix
$\mathbf{B}$ and the matrix with general term
${\rm Bessel}(k,n-k)r^{n-k}$. We have
$$\tilde{\mathbf{B}}_r(n,k)=\frac{n!}{k!}\sum_{j=0}^k\frac{1}{2^j}\binom{n}{k}\frac{r^j}{(n-k-j)!}=\sum_{j=0}^n
\binom{n}{j}\frac{j!r^{j-k}}{(2k-j)!2^{j-k}(j-k)!}.$$ Thus
$$\tilde{\mathbf{B}}_r(n,k)=\binom{n}{k}\sum_{j=0}^n \binom{n-k}{n-j}\frac{k!}{(2k-j)!}\frac{r^{j-k}}{2^{j-k}}$$
and in particular
$$\tilde{\mathbf{B}}_r(2n,n)=\binom{2n}{n}\sum_{j=0}^{2n}\binom{n}{j-n}\frac{n!}{(2n-j)!}\frac{r^{j-n}}{2^{j-n}}.$$ Finally,
$$\tilde{\mathbf{B}}_r(n,k)=\tilde{\mathbf{B}}_r(n-1,k-1)+\tilde{\mathbf{B}}_r(n-1,k)+r(n-1)\tilde{\mathbf{B}}_r(n-2,k-1).$$
\section{Conclusion} The foregoing has shown that the triangles $\mathbf{B}_r$, and more generally $\tilde{\mathbf{B}}_r$, defined in terms of
exponential Riordan arrays, are worthy of further study. Many of the
sequences linked to them have significant combinatorial
interpretations. $\mathbf{B}_{\frac{1}{2}}$ as documented in
\seqnum{A100862} by Deutsch has a clear combinatorial meaning. This
leaves us with the challenge of finding combinatorial
interpretations for the general arrays $\tilde{\mathbf{B}}_r$,
$r\in\mathbf{Z}$.
\section{Acknowledgments}
The author
wishes to thank an anonymous referee for their careful reading of the first version of this paper, and for their pertinent comments and
suggestions.
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\end{thebibliography}

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\noindent 2000 {\it Mathematics Subject Classification}: Primary
11B83; Secondary 05A19, 33C45, 11B37, 11B65.

\noindent \emph{Keywords:} Pascal's triangle, Narayana numbers,
Catalan numbers, Lah numbers, Hermite polynomials, Laguerre
polynomials, Bessel polynomials.

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(Concerned with sequences  \seqnum{A000085}, \seqnum{A000108},
\seqnum{A000262}, \seqnum{A000898}, \seqnum{A001147}, \seqnum{A001263},
\seqnum{A001498},  \seqnum{A001813}, \seqnum{A002720},
\seqnum{A005425}, \seqnum{A007318}, \seqnum{A008297},
\seqnum{A025167}, \seqnum{A025168}, \seqnum{A047974},
\seqnum{A052852}, \seqnum{A100862}, and \seqnum{A102757}.)

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\vspace*{+.1in}
\noindent
Received January 16 2006;
revised version received March 27 2007. 
Published in {\it Journal of Integer Sequences}, March 28 2007.

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\noindent
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